Divergencia Rotacional 90E777

1. **Planteamiento del problema:** Queremos encontrar la divergencia y el rotacional del campo vectorial $$F(x,y,z) = 0\mathbf{i} + \cos(xz)\mathbf{j} - \sin(xy)\mathbf{k}$$. 2. **Fórmulas importantes:** - La divergencia de un campo vectorial $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ es $$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$$ - El rotacional de $\mathbf{F}$ es $$\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}$$ 3. **Identificamos componentes:** $$P = 0, \quad Q = \cos(xz), \quad R = -\sin(xy)$$ 4. **Calculamos la divergencia:** $$\frac{\partial P}{\partial x} = \frac{\partial 0}{\partial x} = 0$$ $$\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y} \cos(xz) = 0 \quad \text{(porque no depende de } y)$$ $$\frac{\partial R}{\partial z} = \frac{\partial}{\partial z} (-\sin(xy)) = 0 \quad \text{(porque no depende de } z)$$ Por lo tanto, $$\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0$$ 5. **Calculamos el rotacional:** - Componente $\mathbf{i}$: $$\frac{\partial R}{\partial y} = \frac{\partial}{\partial y} (-\sin(xy)) = -\cos(xy) \cdot x$$ $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z} \cos(xz) = -\sin(xz) \cdot x$$ Entonces, $$\left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) = -x \cos(xy) - (-x \sin(xz)) = -x \cos(xy) + x \sin(xz)$$ - Componente $\mathbf{j}$: $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial x} (-\sin(xy)) = -\cos(xy) \cdot y$$ $$\frac{\partial P}{\partial z} = \frac{\partial 0}{\partial z} = 0$$ Entonces, $$- \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) = -(-y \cos(xy) - 0) = y \cos(xy)$$ - Componente $\mathbf{k}$: $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \cos(xz) = -\sin(xz) \cdot z$$ $$\frac{\partial P}{\partial y} = \frac{\partial 0}{\partial y} = 0$$ Entonces, $$\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) = -z \sin(xz) - 0 = -z \sin(xz)$$ 6. **Resultado final:** $$\nabla \cdot \mathbf{F} = 0$$ $$\nabla \times \mathbf{F} = \left(-x \cos(xy) + x \sin(xz)\right) \mathbf{i} + y \cos(xy) \mathbf{j} - z \sin(xz) \mathbf{k}$$